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  • Using properties of determinants, prove that <img alt="\begin{vmatrix} 1 & 1 & 1+3x\\ 1+3y & 1 & 1\\ 1& 1+3z &1 \end{vmatrix}= 9\left ( 3xyz+xy+yz+zx \right )" src="https://entrancecorner.codecogs.com/gif.latex?%5Cbegin%7

Using properties of determinants, prove that
\begin{vmatrix} 1 & 1 & 1+3x\\ 1+3y & 1 & 1\\ 1& 1+3z &1 \end{vmatrix}= 9\left ( 3xyz+xy+yz+zx \right )

 

 

 

 
 
 
 
 

Answers (1)

\begin{vmatrix} 1 & 1 & 1+3x\\ 1+3y& 1 & 1\\ 1&1+3z & 1 \end{vmatrix}
Taking x,y and z common from R1,R2 and R3 respectively , we get
= xyz\begin{vmatrix} \frac{1}{x} &\frac{1}{x} & \frac{1}{x}+3\\ \frac{1}{y}+3& \frac{1}{y} & \frac{1}{y}\\ \frac{1}{z}&\frac{1}{z}+3 & \frac{1}{z} \end{vmatrix}
On applying R_{1}\rightarrow R_{1}+R_{2}+R_{3} , we get
= xyz\begin{vmatrix} \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3 &\frac{1}{x}+\frac{1}{y}+\frac{1}{z} +3& \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3\\ \frac{1}{y}+3& \frac{1}{y} & \frac{1}{y}\\ \frac{1}{z}&\frac{1}{z}+3 & \frac{1}{z} \end{vmatrix}
taking common \left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3 \right ) from R1, we get
= xyz\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3 \right )\begin{vmatrix} 1 & 1 &1 \\ \frac{1}{y}+3& \frac{1}{y} & \frac{1}{y} \\ \frac{1}{z}&\frac{1}{z}+3 &\frac{1}{z} \end{vmatrix}

= xyz\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3 \right )\begin{vmatrix} 1 & 0 &0 \\ \frac{1}{y}+3& -3 & -3 \\ \frac{1}{z}& 3 & 0 \end{vmatrix}
On expanding along R1 , we get
= xyz\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3 \right ),\left ( 0+9 \right )
= xyz\left ( \frac{yz+xz+xy+3xyz}{xyz} \right )\left ( 9 \right )
=9\left ( 3xyz+xy+yz+zx \right )
= RHS Hence proved

Posted by

Ravindra Pindel

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